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Given the production function $f(K, L)=\min\{3K,2L\}$, the procedure to find the long-run cost function would be to use the condition: $3K=2L=Y$ where $K=\frac{\overline{Y}}{3}$ and $L=\frac{\overline{Y}}{2}$. $K$ is units of capital; $L$ is units of labor; $Y$ is units of output.

Now, I am being asked to find the short-run cost function where labor is fixed at $L>\frac{\overline{Y}}{2}$. I am not sure how to approach the problem in this case.

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  • $\begingroup$ Since you say labor is fixed, I guess you need the short run cost function where $K$ is the factor that represents the variable cost, while $L$ representing the fixed cost. $\endgroup$ May 23, 2023 at 13:39
  • $\begingroup$ Don't forget you can accept an answer by clicking on the checkmark if you found it helpful! $\endgroup$ May 29, 2023 at 20:20

2 Answers 2

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The long run is defined as the time period in which all the inputs can be changed as per the firm's desire i.e., for a production function of the form $f(K,L)$ both $K$ and $L$ are variable in determining output.

The long-run cost minimization problem is: $$\begin{align} \min_{K,L} \quad &wL+rK \\ \textrm{s.t.} \quad& \min(3K,2L)\geq \bar Y \end{align}$$

The constraint binds at optimum and solving the optimization problem yields conditional input demands $K^d(w,r,\bar Y)=\frac{\bar Y}{3}$ and $L^d(w,r,\bar Y)=\frac{\bar Y}{2}$. Therefore, LR Cost function is: $C(w,r,\bar Y)=\bar Y(\frac{w}{2}+\frac{r}{3})$

The short run is defined as that time period in which one or more inputs are fixed.

The short-run cost minimization problem is: $$\begin{align} \min_{K} \quad & w \bar L+r K \\ \textrm{s.t.} \quad & \min(3K,2 \bar L)\geq\bar Y \\ \end{align}$$

for the constraint $\min(3K,2\bar L)\geq \bar Y$ to hold we need $3K\geq \bar Y$ and $2\bar L\geq \bar Y$

therefore we can rewrite the problem as: $$\begin{align} \min_{K \geq \frac{\bar{Y}}{3}} \quad & w \bar L+r K \\ \end{align}$$

notice that $w\bar L+rK$ is increasing in $K$ therefore in order to solve the above problem we set $K$ to the lowest value it can take.
Thus, $K^d_{SR}(w,r,\bar L, \bar Y)=\frac{\bar Y}{3}$ and consequently SR cost function is: $C_{SR}(w,r,\bar L, \bar Y)=w\bar L+r\frac{\bar Y}{3}$

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  • $\begingroup$ In case we were given that $\bar L<\frac{\bar Y}{2}$ in that case to minimize cost, we would set $3K=2\bar L <\bar Y$ i.e., $K_{SR}^d(w,r,\bar L, \bar Y)=\frac{2\bar L}{3}$. So, I think a more complete solution to the SR cost minimization problem is: $K_{SR}^d(w,r,\bar L, \bar Y)=\begin{cases} \frac{\bar Y}{3} & \text{if } \bar L>\frac{\bar Y}{2} \\ \frac{2 \bar L}{3} & \text{if } \bar L\leq \frac{\bar Y}{2} \end{cases}$ but I'm not sure how correct this is, so decided to include this in the comments $\endgroup$
    – mynameparv
    May 24, 2023 at 10:27
  • $\begingroup$ Thank you! I'm also not sure about the second part but thanks anyway! $\endgroup$
    – Debbie
    May 29, 2023 at 19:21
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Since $L$ is fixed at some $\overline{L} > \frac{\overline{Y}}{2}$, then $\overline{Y} < 2L$.

This implies the $\min$ term equals $3K$.

Therefore, $f(K,\overline{L}) = 3K$.

With this, the problem we want to solve is

$\min_K w\overline{L} + rK$

s.t. $3K \geq \overline{Y}$

Solving for $K$ we get $K \geq \frac{\overline{Y}}{3}$

Clearly the objective function is minimized at equality, so we get $K = \frac{\overline{Y}}{3}$

Plugging into the objective function, the short run minimum cost function is

$C(r,w,\overline{Y};\overline{L}) = w \overline{L} + \frac{r \overline{Y}}{3}$

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  • $\begingroup$ Thanks a lot for your help! $\endgroup$
    – Debbie
    May 29, 2023 at 19:21

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